"FEFF+IFEFFIT" and "cumulant expansion+ratio method" approaches
Dear all, I'm stuck on a peculiar situation and would appreciate any kind of help available. I need to "translate" EXAFS analysis results obtained with IFEFFIT/FEFF into the "language" of the cumulant expansion/ratio method approach. More specifically, I have to relate my mean interatomic distance R obtained with IFEFFIT to the first cumulants (mean interatomic distances) that show up in the ratio method formalism. I'm saying the cumulantS because the ratio method makes a distinction between the so called "effective P(r,lambda)" and "real rho(r)" distributions of interatomic distances, which are related by: P(r,lambda)=rho(r)*[[exp(-2r/lambda)]/r^2] . For the second cumulant (Debye-Waller factor or sigma^2) and higher terms, the difference between "effective" and "real" values is not significant unless the disorder in the sample is really big. But for the first cumulant it is significant (at least at not very low temperatures), being the "real" first cumulant bigger than the "effective" one by a term like [(2*sigma^2)/r]*[1+(r/lambda)]. My dilemma is: how my mean interatomic distance R from IFEFFIT relates to the "effective" and "real" first cumulants? Should it be the same as one of them? Which one? Or it doesn't correspond exactly to any of them? Any comments will be welcome... Regards, Leandro
Hi Leandro, Sorry for the delay. XAFS 13 was last week, and I'm still catching up!! Ifeffit reports (as well as it can) cumulants of the "real" distribution function, including corrections for the mean-free-path, and the 1/R^2 term in the EXAFS equation. There should be no need to convert back to any sort of 'effective pair distribution' as might be found be "other methods", including log-ratio methods, etc. The burden should be on those methods to correct their effective distributions to the real ones. So, my immediate reaction would be to not try to do this. Assuming that you still want to, I think it's not a trivial conversion because a) lambda is k-dependent, b) Feffit and Ifeffit use p = k+ Sigma(k) + i/lambda(k) as the conjugate variable to R, and so the expansion coefficient for the cumulant expansion. Taking lambda(k) into account is part of the problem -- so is the Sigma(k). That additional self-energy term includes an estimate of the difference between E_fermi (where k=0) and E_0 (where p=0). You can see this in the feff.dat files (the column k and Re[p] are slightly different). (And it should be that Feff8 is better at this than Feff6, but the jury's still out for how big that effect is). In ifeffit, you can see this with > path(1, feff0001.dat) > get_path(1, group=f1, do_arrays) > plot f1.k, f1.rep But that's all to say that the conversion is not that easy. Perhaps you could explain what the problem you're trying to solve and why you might want the cumulants of an effective potential? --Matt
Hi Matt, thanks a lot for the reply. I actually don't want to and don't have to convert my Ifeffit results to any kind of effective pair distribution function. I understand that it is not a trivial conversion and some approximations would be necessary. I just needed to be sure that by using Ifeffit I get the "real" cumulants, instead of the so-called "effective" ones, and no "post-analysis" corrections are necessary to the Ifeffit results. Knowing that is enough for me to go on and compare my results to some others obtained by different methods. And the results and talking about are the thermal evolution of the cumulants for crystalline Ge and Ge nanocrystals. Cheers, Leandro Matt Newville wrote:
Hi Leandro,
Sorry for the delay. XAFS 13 was last week, and I'm still catching up!!
Ifeffit reports (as well as it can) cumulants of the "real" distribution function, including corrections for the mean-free-path, and the 1/R^2 term in the EXAFS equation.
There should be no need to convert back to any sort of 'effective pair distribution' as might be found be "other methods", including log-ratio methods, etc. The burden should be on those methods to correct their effective distributions to the real ones.
So, my immediate reaction would be to not try to do this. Assuming that you still want to, I think it's not a trivial conversion because a) lambda is k-dependent, b) Feffit and Ifeffit use p = k+ Sigma(k) + i/lambda(k) as the conjugate variable to R, and so the expansion coefficient for the cumulant expansion. Taking lambda(k) into account is part of the problem -- so is the Sigma(k).
That additional self-energy term includes an estimate of the difference between E_fermi (where k=0) and E_0 (where p=0). You can see this in the feff.dat files (the column k and Re[p] are slightly different). (And it should be that Feff8 is better at this than Feff6, but the jury's still out for how big that effect is). In ifeffit, you can see this with
path(1, feff0001.dat) get_path(1, group=f1, do_arrays) plot f1.k, f1.rep
But that's all to say that the conversion is not that easy.
Perhaps you could explain what the problem you're trying to solve and why you might want the cumulants of an effective potential?
--Matt _______________________________________________ Ifeffit mailing list Ifeffit@millenia.cars.aps.anl.gov http://millenia.cars.aps.anl.gov/mailman/listinfo/ifeffit
participants (3)
-
Leandro Araujo -
Leandro Langie Araujo
-
Matt Newville